functions

- mathmath333

functions

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- mathmath333

\(\large \color{black}{\begin{align} &\normalsize \text{Find the maximim value of the function}\hspace{.33em}\\~\\
& \dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\
\end{align}}\)

- SolomonZelman

interval given ?

- SolomonZelman

with out interval this maximum value diverges to infinity.

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## More answers

- mathmath333

this question means that i have to minimize
\(\large \color{black}{\begin{align}
x^2-3x+2 \hspace{.33em}\\~\\
\end{align}}\)
right ?

- SolomonZelman

should be, or els... i mean there is not maximum value here, if you are taking over \((-\infty,~+\infty)\)

- SolomonZelman

wait

- SolomonZelman

1/that ? or, just that ?

- mathmath333

lol its
\(\large \color{black}{\begin{align}
\dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\
\end{align}}\)

- SolomonZelman

\(\large\color{black}{ \displaystyle f(x)=\frac{1}{x^2-3x+2} }\)
and you want to find absolute minimum, right?

- SolomonZelman

well, I can tell you that the limit as x approaches \(\pm\)infinity is going to be 0.

- SolomonZelman

we can plot, I am guessing it is something like
|dw:1435010239497:dw|

- mathmath333

i want to find the maximum value of this
\(\large \color{black}{\begin{align}
\dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\
\end{align}}\)

- SolomonZelman

none

- SolomonZelman

it is +infinity

- mathmath333

but how ?

- SolomonZelman

https://www.desmos.com/calculator

- mathmath333

how i prove that with algebra

- SolomonZelman

even the minimum doesn't exist.

- SolomonZelman

with algebra, even with no calc ?

- mathmath333

idk calculus

- SolomonZelman

ic

- SolomonZelman

lets c

- SolomonZelman

\(\large\color{black}{ \displaystyle f(x)=\frac{1}{x^2-3x+2} }\)
it has vertical asymptotes, can you find them for me?

- mathmath333

\(\large \color{black}{\begin{align}
x=2,\ 1 \hspace{.33em}\\~\\
\end{align}}\)

- SolomonZelman

yes

- SolomonZelman

and, the best thing I can see here is to use the idea of a limit to see where range tends to the close we get to x=1 and x=2.
I don't really know a purely algebraic proof for why this function has no max or min.

- xapproachesinfinity

that has a max actually

- SolomonZelman

yes?

- SolomonZelman

https://www.desmos.com/calculator
keep scrollin up

- SolomonZelman

on a certain interval, no doubt (unless it has a domain gap)

- xapproachesinfinity

i think it attains one max
does not perhaps required any domain to get it or i think

- xapproachesinfinity

your graph does not show up
see here
https://www.desmos.com/calculator

- SolomonZelman

i don't think there is a max. that is due to an asymptote - two of them, there.

- SolomonZelman

we can find the limit from the right and from the left as x approaches 1 and 2.

- xapproachesinfinity

http://prntscr.com/7k4q1l
here see that bottom part of the graph

- xapproachesinfinity

with calculus we can prove it has a max value
but @mathmath333 is looking for a different way

- SolomonZelman

this graph goes further down and up if you scroll on desmos, but we can go ahead and take any limit that is closed from either side of any of the asymptotes you wish to choose.....

- xapproachesinfinity

i mean here local max not abolute of course

- mathmath333

how to find that local max

- SolomonZelman

oh, but the question is asking for absolute doesn't it?

- SolomonZelman

that is what i would think of course

- xapproachesinfinity

first let's write that as 1/(x-1)(x-2)

- xapproachesinfinity

i had thought of absolute too, just not sure about it!

- mathmath333

yes the question is asking absolute, i just asked that of curiosity

- SolomonZelman

this function is pretty weird. I wonder what would it model

- SolomonZelman

i mean in real world

- xapproachesinfinity

you want to then prove there is no max algebraically

- mathmath333

yes ^

- SolomonZelman

would that include using ideas from calc but not its techniques?

- SolomonZelman

I am stubborn on the idea of a limit. A limit would simply show that as it approaches __ it can be infinitely large/small

- mathmath333

ok i will try to bear with calculus

- SolomonZelman

\(\large\color{black}{ \displaystyle \frac{1 }{x^2-3x+2} }\)
we need to do only 1 asymptote for this

- SolomonZelman

\(\large\color{black}{ \displaystyle \frac{1 }{x^2-3x+2} }\)
lets plug in values that are less than, but close to 1

- SolomonZelman

\(\large\color{black}{ \displaystyle f(0.9)=? }\)

- SolomonZelman

= about 9

- SolomonZelman

f(0.99)=99

- SolomonZelman

try plugging it yourself for close values to 1

- SolomonZelman

these values are called approaching "from the left" because they are a buit less than 1

- SolomonZelman

now, try to do
f(1.5)
f(1.1)
f(1.01)
f(1.001)
and you will see where it goes "from the right" - for values that are greater than 1

- SolomonZelman

i got to go

- SolomonZelman

sorry

- xapproachesinfinity

suppose there exist a max
max(1/(x^2-3x+2)=c
1/x^2-3x+21/c=l
x^2-3x+2-l>0

- xapproachesinfinity

just some random stuff lol

- xapproachesinfinity

coming back..

- mathmath333

ok i need to go please comment if u found something

- xapproachesinfinity

let's first consider the domain
\(D_f: \{ x\in \mathbb{R}: x\ne1, x\ne 2 \}\)

- xapproachesinfinity

that is (-oo, 1)u(1, 2)u(2,oo)

- xapproachesinfinity

we look at the sign of y when x<1 if y does not change the sign in that interval then
i must not have a max
seems that 1/(x-2)(x-1)>0 for x<1
same reasoning for x>2 ===> y>0
so only btw 1 and 2 left

- xapproachesinfinity

from the graph we knew for 1

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